Solve for $q$, $ \dfrac{5}{4q^2} = \dfrac{10}{4q^2} + \dfrac{q - 8}{2q^2} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4q^2$ $4q^2$ and $2q^2$ The common denominator is $4q^2$ The denominator of the first term is already $4q^2$ , so we don't need to change it. The denominator of the second term is already $4q^2$ , so we don't need to change it. To get $4q^2$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{q - 8}{2q^2} \times \dfrac{2}{2} = \dfrac{2q - 16}{4q^2} $ This give us: $ \dfrac{5}{4q^2} = \dfrac{10}{4q^2} + \dfrac{2q - 16}{4q^2} $ If we multiply both sides of the equation by $4q^2$ , we get: $ 5 = 10 + 2q - 16$ $ 5 = 2q - 6$ $ 11 = 2q $ $ q = \dfrac{11}{2}$